Evaluate $\displaystyle 2\pi \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \tan x + \sqrt{1 + \sec ^4 x} dx$
i know the final anwser i need to know how to do it. its not for homework or a test i just want to know how to do it
Evaluate $\displaystyle 2\pi \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \tan x + \sqrt{1 + \sec ^4 x} dx$
i know the final anwser i need to know how to do it. its not for homework or a test i just want to know how to do it
lets concentrate on $\displaystyle \int \tan x \sqrt{1 + \sec^4 x}~dx$
Note that this is the same as $\displaystyle \int \frac {\sec^4x \tan x \sqrt{1 + \sec^4 x}}{\sec^4 x}~dx$ ........i multiplied by $\displaystyle \frac {\sec^4 x}{\sec^4 x}$
Now, let $\displaystyle u^2 = 1 + \sec^4 x$
$\displaystyle \Rightarrow 2u~du = 4 \sec^4 x \tan x~dx$
and $\displaystyle \sec^4 x = u^2 - 1$
So our integral becomes:
$\displaystyle \frac 12 \int \frac {u^2}{u^2 - 1}~du$
which is relatively easy to deal with
indeed haha! actually, this problem came from http://www.mathhelpforum.com/math-he...plication.html
and look at my long solution, i thought many things at the same time!
well, look at that. i never remembered seeing this integral before. how did you?!
it was probably a mistake that i stumbled on it
i just wanted to get rid of the square root. so i made what's underneath it $\displaystyle u^2$. then i realized that i would need a $\displaystyle sec^4 x$ that wasn't there, so i put it there. wonder how it came to me
haha, no, and for good reason